Equality in prime number inequality Is $3 = 2 + 1$ the only case when equality holds in the following :
$$ p_{n} \leq p_{1}p_{2} \cdots p_{n-1} + 1$$
where $p_{r}$ denotes the $r^{th}$ prime number.
 A: It is the only case. 
We have $p_n < p_1\dots p_{n-1}$ for $n \geq 3$. We can prove this by induction. 
For $n=3$ we have $5 < 2\times 3 = 6$. Now suppose $p_{n-1} < p_1\dots p_{n-2}$. By Bertrand's postulate, there is a prime $p^*$ between $p_1\dots p_{n-2}$ and $2p_1\dots p_{n-2}$. As $p_1, \dots, p_{n-2} < p_1\dots p_{n-2}$, and by the induction hypothesis $p_{n-1} < p_1\dots p_{n-2}$, we have $p^* \neq p_1, \dots, p_{n-1}$, so $p^* \geq p_n$. Therefore $p_n \leq p^* < 2p_1\dots p_{n-2} < p_1\dots p_{n-1}$.
A: Let $P=p_1\cdot p_2\cdots p_n$ be the product of the first $n$ primes with $p_n>3$.  
Then $P$ can be written as the sum of two integers which are coprime to $P$  say $P=a+b$.
Suppose now that $p|a$ and $q|b$ for some primes $p,q$.We suppose that $p<q$.
Because $\gcd(a,P)=\gcd(b,P)=1$ we can see that $p>p_n$ so $p\geq p_{n+1}$ and $q\geq p_{n+2}$ .  
But $a+b\geq p+q\geq p_{n+1}+p_{n+2}\geq 2\cdot p_{n+1}$ .
So,we have proved that $2\cdot p_{n+1} \leq a+b= P$ and so,$p_{n+1}\leq P/2<P-1$.
So yes equality holds only for $3$.
A: Well, let $q = p_1 *p_2 * ...*p_{n-1} +1$
Then $q$ cannot be divided by $p_i$ for $i\leq (n-1)$ and $q$ is not equal to any of these $p_i$.
Thus either $q$ is prime or has a prime factor greater than $p_{n-1}$. 
If $q$ is prime, obviously $q \geq p_n$. 
If $q$ has a prime factor greater than $p_{n-1}$, then this prime factor must be $\geq p_n$. Thus $q \geq p_n$. 
Update: 
Actually, Michael Albanese's answer shows, that according to Bertrands postulate: pn < 2 * pn-1. and therefore pn < p1 * ...* pn-2 * pn-1 if 2< p1*...*pn-2 (#) This obviously is the case, as soon as pn-2 >2, i.e. as soon as on the right side of the equation (#) the product is at least 2*3 = 6 (and it will always remain > 2, of course)
